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25-Nav-B1 Applied Thermodynamics and Heat Transfer (25-Mec-A1) · December 2013

Question 2 of 6: Stepped Shaft — Deflection and Critical Speed

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2013 · 98-Mar-B1 Advanced Machine Design. Open-book, 3 hours, 100 marks. The paper requires all of Part I (Problems 1 and 2) plus any three of the four Part II problems (3–6). All six problems are solved in full below.

Reference texts. R. G. Budynas & J. K. Nisbett, Shigley’s Mechanical Engineering Design, 10th ed. (shafts & critical speed §7; bolted joints §8; lubrication & journal bearings §12; brakes §16); R. C. Juvinall & K. M. Marshek, Fundamentals of Machine Component Design (bearings, brakes, impact); R. C. Hibbeler, Mechanics of Materials (bending, impact loading).

Check: this paper, although listed under Applied Thermodynamics and Heat Transfer, is headed “98-Mar-B1” with printed title “Advanced Machine Design”; it is a general mechanical machine-design paper — stress/yield criteria, shaft deflection/critical speed, bolted-joint preload, journal-bearing sizing, impact loading, and drum brakes — with zero thermodynamics or heat-transfer content, and it is solved as the exam actually printed.

Question 2: Stepped Shaft — Deflection and Critical Speed (30 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A simply-supported (bearing-supported) stepped steel shaft, span 800 mm between bearings at $A$ and $F$, carrying two transverse loads.

Given data (steel, $E = 200$ GPa)
QuantityValue
Bearing span $A\to F$800 mm
Load $P_1$ at $B$ ($x=200$ mm)10 kN (down)
Load $P_2$ at $D$ ($x=500$ mm)20 kN (down)
Diameter $A\to C$ and $E\to F$$d_1 = 100$ mm
Diameter $C\to E$ (middle)$d_2 = 125$ mm
Segment lengths (A-B-C-D-E-F)200 / 100 / 200 / 200 / 100 mm
P1 = 10 kNP2 = 20 kNABCDEFd = 100d = 125800 mm (segments: 200 / 100 / 200 / 200 / 100)
Stepped shaft: bearings $R_A$, $R_F$; loads $P_1=10$ kN at $B$ and $P_2=20$ kN at $D$; diameters 100 mm (outer) and 125 mm (centre).

Find. The maximum static deflection and where it occurs, the fundamental critical (whirling) speed, and whether running at 1000 rpm is safe.

Approach. Find the bearing reactions and bending-moment diagram, integrate $M(x)/EI(x)$ twice along the stepped section to get the elastic curve (the step in $I$ is carried explicitly), then apply Rayleigh’s energy method to the two rotating loads to obtain the first critical speed and compare it with the operating speed.

  1. Bearing reactions. Taking moments about $A$, $R_F\,(800) = P_1(200)+P_2(500) = 10(200)+20(500) = 12{,}000$ kN·mm, so $R_F = 15$ kN and $R_A = 30-15 = 15$ kN by vertical equilibrium.
  2. Bending-moment diagram. With $x$ from $A$, $M = R_Ax$ up to $B$, then $M = R_Ax-P_1(x-200)$, then $M = R_Ax-P_1(x-200)-P_2(x-500)$. The peak is at $D$: $$M_{\max} = 15(500)-10(300) = \boxed{4.5\ \text{kN}\cdot\text{m}\ \text{at }D}.$$ The value at $B$ is $3.0$ kN·m, and $M=0$ at both bearings, as required.
  3. Section stiffness. The two second moments are $I_1 = \dfrac{\pi d_1^{4}}{64} = \dfrac{\pi (100)^4}{64} = 4.909\times10^{6}\ \text{mm}^4$ and $I_2 = \dfrac{\pi (125)^4}{64} = 1.198\times10^{7}\ \text{mm}^4$, giving flexural rigidities $EI_1 = 9.82\times10^{11}$ and $EI_2 = 2.40\times10^{12}\ \text{N}\cdot\text{mm}^2$.
  4. Elastic curve (stepped-$EI$ double integration). Integrating the curvature $y''(x)=M(x)/[E\,I(x)]$ twice with $y(0)=y(800)=0$ (numerically, so the abrupt change of $I$ at $C$ and $E$ is handled exactly) gives the deflected shape. The largest deflection is $$\boxed{y_{\max} = 0.152\ \text{mm at } x \approx 356\ \text{mm}},$$ i.e. within the stiff central section, between the two loads. The deflections under the loads themselves are $y_B = 0.124$ mm and $y_D = 0.134$ mm.
  5. Fundamental critical speed (Rayleigh). Treating $P_1,P_2$ as the weights of the mounted rotating masses and using their static deflections, $$\omega_n = \sqrt{\dfrac{g\,\sum W_i y_i}{\sum W_i y_i^{2}}} = \sqrt{\dfrac{9810\,[10(0.124)+20(0.134)]}{10(0.124)^2+20(0.134)^2}} = 274\ \text{rad/s},$$ (with $y$ in mm, $g=9810\ \text{mm/s}^2$). Converting, $$\boxed{N_{cr} = \dfrac{60\,\omega_n}{2\pi} \approx 2610\ \text{rpm}}.$$ Rayleigh’s method slightly over-estimates, so the true first critical speed is a little below this.
  6. Safety at 1000 rpm. The operating speed is $1000/2610 = 0.38$ of the critical speed. Good practice keeps the running speed below about $0.75\,N_{cr}$ (or above $1.4\,N_{cr}$) to stay clear of the whirl resonance. At 38% of critical the shaft runs well below its first whirl, so yes, 1000 rpm is safe.
Problem 2 — results
QuantityResult
Reactions $R_A,\,R_F$15 kN, 15 kN
Maximum bending moment4.5 kN·m at $D$
Maximum deflection0.152 mm at $x\approx356$ mm
Fundamental critical speed$\approx 2610$ rpm
Operating 1000 rpm0.38 $N_{cr}$ → safe